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A cyclotron is used to produce a beam of high-energy deuterons that then collide with a target to produce radioactive isotopes f

serg [7]

Answer:

19080667.0818 m/s

0.637294 m

$2.1875\times 10^{15}$

Explanation:

m = Mass of deuterons = $3.34\times 10^{-27}\ kg$

v = Velocity

K = Kinetic energy = 3.8 MeV

d = Diameter

B = Magnetic field = 1.25 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

t = Time = 1 s

i = Current = 350 μA

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2K}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 3.8\times 10^6\times 1.6\times 10^{-19}}{3.34\times 10^{-27}}}\\\Rightarrow v=19080667.0818\ m/s$

The speed of the deuterons when they exit is 19080667.0818 m/s

In this system the centripetal and magnetic force will balance each other

$\dfrac{mv^2}{r}=qvB\\\Rightarrow \dfrac{mv^2}{\dfrac{d}{2}}=qvB\\\Rightarrow d=\dfrac{2mv}{qB}\\\Rightarrow d=\dfrac{2\times 3.34\times 10^{-27}\times 19080667.0818}{1.6\times 10^{-19}\times 1.25}\\\Rightarrow d=0.637294\ m$

The diameter is 0.637294 m

Current is given by

$i=\dfrac{nq}{t}\\\Rightarrow n=\dfrac{it}{q}\\\Rightarrow n=\dfrac{350\times 10^{-6}\times 1}{1.6\times 10^{-19}}\\\Rightarrow n=2.1875\times 10^{15}$

The number of deuterons is $2.1875\times 10^{15}$

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A machine does 50 J of Work on a 10 Kg Object. What is the Change in the Objects Total Energy?

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The change in the total energy of the object is zero (0).

The given parameters:

work done by the machine, W = 50 J

mass of the object, m = 10 kg

To find:

the change in the total energy of the object

The change in the total energy of the object is the difference between the objects initial energy due to its position and the work done on the object.

Based on work energy-theory, the work done on the object is equal to the energy of the object.

The energy of the object = work-done on the object

The change in total energy = 50 J - 50 J = 0

Thus, the change in the total energy of the object is zero (0).

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In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x=

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In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x=20.0 cm to x=60.0cm.

<h3>Where is the electric potential, when the particle moved?</h3>

The charge field system's electric potential energy rose. The particle experiences an electric force that is directed against the x-axis. It is pushed uphill by an outside force, which raises the potential energy.

When a charge to be moved against an applied electric field, electric potential energy is needed. A charge must be moved through a stronger electric field with more energy than it would require to carry it via a weaker electric field.

In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x=20.0 cm to x=60.0cm.

The electric potential energy of the charge field system:

(a) increase
(b) remain constant
(c) decrease
(d) change unpredictably

The correct option is a).

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